If the radius of the earth were to shrink by one percent its mass remaining the same, the acceleration due to greavity on the earth's surface would

Options

A. decreases
B. remain unchanged
C. increases
D. it will depend on the mass of earth

Answer

To solve the problem, we will use the formula for acceleration due to gravity at the surface of the Earth, which is given by:

\[ g = \frac{G \cdot M}{R^2} \]

where:
- \( g \) is the acceleration due to gravity,
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

### Step 1: Understand the change in radius
Given that the radius of the Earth shrinks by 1%, we can express the new radius \( R' \) as:

\[ R' = R - 0.01R = 0.99R \]

### Step 2: Substitute the new radius into the formula
The new acceleration due to gravity \( g' \) can be expressed using the new radius:

\[ g' = \frac{G \cdot M}{(R')^2} = \frac{G \cdot M}{(0.99R)^2} \]

### Step 3: Simplify the expression
Now, we can simplify \( g' \):

\[ g' = \frac{G \cdot M}{(0.99^2 \cdot R^2)} = \frac{G \cdot M}{0.9801 \cdot R^2} \]

### Step 4: Relate \( g' \) to the original \( g \)
We know that the original acceleration due to gravity \( g \) is:

\[ g = \frac{G \cdot M}{R^2} \]

Now, we can express \( g' \) in terms of \( g \):

\[ g' = \frac{g}{0.9801} \]

### Step 5: Analyze the result
Since \( 0.9801 < 1 \), it follows that:

\[ g' > g \]

This means that the acceleration due to gravity increases when the radius of the Earth shrinks by 1% while the mass remains constant.

### Conclusion
Thus, the acceleration due to gravity on the Earth's surface would increase.

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